Problem: Let $g(x)=\begin{cases} \dfrac{\sqrt{x+20}-4}{x+4}&\text{for }x\geq -20, x\neq -4 \\\\ k&\text{for }x=-4 \end{cases}$ $g$ is continuous for all $x>-20$. What is the value of $k$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{1}{8}$ (Choice B) B $\dfrac{1}{4}$ (Choice C) C $-\dfrac{1}{4}$ (Choice D) D $-\dfrac{1}{8}$
Explanation: $\dfrac{\sqrt{x+20}-4}{x+4}$ is continuous for all $x>-20$ other than $x=-4$, which means $g$ is continuous for all $x>-20$ other than $x=-4$. In order for $g$ to also be continuous at $x=-4$, the following equality must hold: $\lim_{x\to -4}g(x)=g(-4)$ Since $g(-4)=k$, we will obtain the above equality by letting $k=\lim_{x\to -4}g(x)$. So let's find $\lim_{x\to -4}g(x)$, come on! $\begin{aligned} &\phantom{=}\lim_{x\to -4}g(x) \\\\ &=\lim_{x\to -4}\dfrac{\sqrt{x+20}-4}{x+4} \gray{\text{This is the rule for }x\neq -4} \\\\ &=\lim_{x\to -4}\dfrac{\sqrt{x+20}-4}{x+4}\cdot\dfrac{\sqrt{x+20}+4}{\sqrt{x+20}+4} \gray{\text{Rationalize}} \\\\ &=\lim_{x\to -4}\dfrac{x+20-4^2}{(x+4)(\sqrt{x+20}+4)} \gray{\text{Simplify}} \\\\ &=\lim_{x\to -4}\dfrac{\cancel{x+4}}{\cancel{(x+4)}(\sqrt{x+20}+4)} \gray{\text{Cancel common factors}} \\\\ &=\lim_{x\to -4}\dfrac{1}{(\sqrt{x+20}+4)} \\\\ &\text{(This is allowed because }x\neq -4) \\\\ &=\dfrac{1}{\sqrt{-4+20}+4} \gray{\text{Direct substitution}} \\\\ &=\dfrac{1}{8} \end{aligned}$ We obtained that if we set $k=\dfrac{1}{8}$, then $\lim_{x\to -4}g(x)=g(-4)$, which makes $g$ continuous at $x=-4$. Since we already saw that $g$ is continuous for any other $x>-20$, we can determine that it's continuous for all $x>-20$. In conclusion, $k=\dfrac{1}{8}$.